Problem on adjoints of Linear Operators.

Question

Could you help me check my solution to the question-

Suppose $T\in L(V)$ and $U$ is a subspace of $V$, show that $U$ is invariant under $T$ if $U^\perp$ is invariant under $T^*$.

Solution: Let $U$ be invariant under $T$ and $z$ be a vector in $U^\perp$, then $T(x)= y$, $x \in U, y\in U,$ now $$ \langle T(x),z\rangle =0 =\langle x, T^*(z)\rangle $$ I.e $\langle x, T^*(z)\rangle=0,$ now x is not zero, then there are two possibilities, 1)$T^*(z)=0$ , 2) $T^*(z) \in U^\perp$. 1) will hold if $\langle x, T^*(z)\rangle =0$ for all $x\in V$ since$\langle x, 0\rangle=0$ for all $x \in V$, but if $x \in U^\perp$, then $\langle x, T^*(z) \rangle \neq 0$, hence 1) can't be true and therefore 2) has to be true and thus $U^\perp$ is invariant under $T^*$.

Answer 1

I can't really follow your proposed solution, I suggest that you make it into points, but here is how I would prove this

• assume that $U$ is not invariant under $T$ then
• for each vector $v$ in $U$, $Tv = u + u^\perp $ where $u \in U, u^\perp \in U^\perp $
• let $T^{\dagger}$ be the adjoint operator of $T$ then
  • $T^{\dagger}Tv = T^{\dagger}u + T^{\dagger}u^\perp\\ v = T^{\dagger}u + T^{\dagger}u^\perp$
  • since $U^\perp$ is inveriant under $T$ we get
    • $T^\dagger u$ has to be a vector in $U$, because no vector in $U^\perp$ maps to a vector in $U$.
    • there is a vector $u' \in U^\perp$ such that $u' = T^\dagger u^\perp$
• from this $v$ has to have a component in $U^\perp$ which is a contradiction